Question: You have found the following ages (in years) of all 6 turtles at your local zoo: $ 20,\enspace 32,\enspace 22,\enspace 86,\enspace 68,\enspace 28$ What is the average age of the turtles at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{20 + 32 + 22 + 86 + 68 + 28}{{6}} = {42.7\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $20$ years $-22.7$ years $515.29$ years $^2$ $32$ years $-10.7$ years $114.49$ years $^2$ $22$ years $-20.7$ years $428.49$ years $^2$ $86$ years $43.3$ years $1874.89$ years $^2$ $68$ years $25.3$ years $640.09$ years $^2$ $28$ years $-14.7$ years $216.09$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{515.29} + {114.49} + {428.49} + {1874.89} + {640.09} + {216.09}} {{6}} $ $ {\sigma^2} = \dfrac{{3789.34}}{{6}} = {631.56\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{631.56\text{ years}^2}} = {25.1\text{ years}} $ The average turtle at the zoo is 42.7 years old. There is a standard deviation of 25.1 years.